Question:
The S-parameters of a two port network is given as \[ [S] = \begin{bmatrix} S_{11} & S_{12} \\[2pt] S_{21} & S_{22} \end{bmatrix} \] with reference to \(Z_0\). Two lossless transmission line sections of electrical lengths \(\theta_1 = \beta l_1\) and \(\theta_2 = \beta l_2\) are added to the input and output ports for measurement purposes, respectively. The S-parameters \([S']\) of the resultant two port network is 
The S-parameters of a two port network is given as \[ [S] = \begin{bmatrix} S_{11} & S_{12} \\[2pt] S_{21} & S_{22} \end{bmatrix} \] with reference to \(Z_0\). Two lossless transmission line sections of electrical lengths \(\theta_1 = \beta l_1\) and \(\theta_2 = \beta l_2\) are added to the input and output ports for measurement purposes, respectively. The S-parameters \([S']\) of the resultant two port network is
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Adding matched, lossless lines only {shifts phase}. For a two-port, $S'_{ij}=S_{ij}\,e^{-j(\theta_i+\theta_j)}$—double the phase for reflections ($i=j$), sum for transmissions.
Updated On: Feb 12, 2026
- $\displaystyle \begin{bmatrix} S_{11}e^{-j2\theta_1} & S_{12}e^{-j(\theta_1+\theta_2)} \\ S_{21}e^{-j(\theta_1+\theta_2)} & S_{22}e^{-j2\theta_2} \end{bmatrix} $
- $\displaystyle \begin{bmatrix} S_{11}e^{j2\theta_1} & S_{12}e^{-j(\theta_1+\theta_2)} \\ S_{21}e^{-j(\theta_1+\theta_2)} & S_{22}e^{j2\theta_2} \end{bmatrix} $
- $\displaystyle \begin{bmatrix} S_{11}e^{j2\theta_1} & S_{12}e^{j(\theta_1+\theta_2)} \\ S_{21}e^{j(\theta_1+\theta_2)} & S_{22}e^{j2\theta_2} \end{bmatrix} $
- \; $\displaystyle \begin{bmatrix} S_{11}e^{-j2\theta_1} & S_{12}e^{j(\theta_1+\theta_2)}
[4pt] S_{21}e^{j(\theta_1+\theta_2)} & S_{22}e^{-j2\theta_2} \end{bmatrix}$
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The Correct Option is A
Solution and Explanation
Step 1: Reference-plane shift for scattering variables.
Adding a matched, lossless line of length \(l_k\) at port \(k\) multiplies both forward and reverse waves at that port by \(e^{-j\theta_k}\), where \(\theta_k=\beta l_k\).
With \(a,b\) the old waves and \(a',b'\) the new ones:
\[ a = D\,a',\qquad b' = D\,b,\qquad D=\operatorname{diag}(e^{-j\theta_1},e^{-j\theta_2}). \]
Step 2: Transform the S-matrix.
Since \(b=Sa\), we get
\[ b' = D\,b = D\,S\,a = D\,S\,D\,a' \;\Rightarrow\; S' = D\,S\,D. \]
Step 3: Entry-wise result.
Thus \(S'_{ij}=S_{ij}\,e^{-j(\theta_i+\theta_j)}\), giving
\[ [S']= \begin{bmatrix} S_{11}e^{-j2\theta_1} & S_{12}e^{-j(\theta_1+\theta_2)} \\[6pt] S_{21}e^{-j(\theta_1+\theta_2)} & S_{22}e^{-j2\theta_2} \end{bmatrix}. \]
\[ \boxed{\text{Option (A)}} \]
Adding a matched, lossless line of length \(l_k\) at port \(k\) multiplies both forward and reverse waves at that port by \(e^{-j\theta_k}\), where \(\theta_k=\beta l_k\).
With \(a,b\) the old waves and \(a',b'\) the new ones:
\[ a = D\,a',\qquad b' = D\,b,\qquad D=\operatorname{diag}(e^{-j\theta_1},e^{-j\theta_2}). \]
Step 2: Transform the S-matrix.
Since \(b=Sa\), we get
\[ b' = D\,b = D\,S\,a = D\,S\,D\,a' \;\Rightarrow\; S' = D\,S\,D. \]
Step 3: Entry-wise result.
Thus \(S'_{ij}=S_{ij}\,e^{-j(\theta_i+\theta_j)}\), giving
\[ [S']= \begin{bmatrix} S_{11}e^{-j2\theta_1} & S_{12}e^{-j(\theta_1+\theta_2)} \\[6pt] S_{21}e^{-j(\theta_1+\theta_2)} & S_{22}e^{-j2\theta_2} \end{bmatrix}. \]
\[ \boxed{\text{Option (A)}} \]
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